Question: Simplify the following expression and state the condition under which the simplification is valid: $k = \dfrac{q^2 - 4q - 21}{q^2 - 6q - 7}$
Solution: First factor the expressions in the numerator and denominator. $ \dfrac{q^2 - 4q - 21}{q^2 - 6q - 7} = \dfrac{(q + 3)(q - 7)}{(q + 1)(q - 7)} $ Notice that the term $(q - 7)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(q - 7)$ gives: $k = \dfrac{q + 3}{q + 1}$ Since we divided by $(q - 7)$, $q \neq 7$. $k = \dfrac{q + 3}{q + 1}; \space q \neq 7$